Question

If a simple harmonic motion is represented by $\frac{d^{2}x}{d{t}^2}+\alpha x=0$ its time period is

• A. $\frac{2\pi }{\alpha }$
  
• B. $\frac{2\pi }{\sqrt{\alpha }}$
  
• C. $2\pi \alpha$
  
  
• D. $2\pi \sqrt{\alpha }$
  

Answer 1

The correct option is B

$\frac{2\pi }{\sqrt{\alpha }}$

Given that, $\frac{d^{2}x}{d{t}^2}+\alpha x=0\Rightarrow \frac{d^{2}x}{d{t}^2}=-\alpha x=-{\omega }^{2}x\Rightarrow \omega =\sqrt{\alpha }\therefore T=\frac{2\pi }{\omega }=\frac{2\pi }{\sqrt{\alpha }}$